Two opposite and equal charges of 4 times 10^ | vedclass

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(D) Given:Charge $q = 4 	imes 10^{-8} \ C$Separation distance $2a = 2 	imes 10^{-2} \ cm = 2 	imes 10^{-4} \ m$Electric field $E = 4 	imes 10^8 \

Vedclass · Accepted Answer

$32 	imes 10^{-4} \ Nm, 64 	imes 10^{-4} \ J$

Vedclass · Answer

(D) Given:Charge $q = 4 	imes 10^{-8} \ C$Separation distance $2a = 2 	imes 10^{-2} \ cm = 2 	imes 10^{-4} \ m$Electric field $E = 4 	imes 10^8 \ N/C$$1$. Dipole moment $p = q 	imes (2a) = (4 	imes 10^{-8} \ C) 	imes (2 	imes 10^{-4} \ m) = 8 	imes 10^{-12} \ Cm$$2$. Maximum torque $	au_{max} = pE \sin(90^{\circ}) = pE$$	au_{max} = (8 	imes 10^{-12} \ Cm) 	imes (4 	imes 10^8 \ N/C) = 32 	imes 10^{-4} \ Nm$$3$. Work done to rotate by $180^{\circ}$ (from $0^{\circ}$ to $180^{\circ}$):$W = pE(\cos 	heta_1 - \cos 	heta_2) = pE(\cos 0^{\circ} - \cos 180^{\circ})$$W = (32 	imes 10^{-4} \ Nm) 	imes (1 - (-1)) = 32 	imes 10^{-4} 	imes 2 = 64 	imes 10^{-4} \ J$

Two opposite and equal charges of $4 \times 10^{-8} \ C$ are placed at a distance of $2 \times 10^{-2} \ cm$ to form a dipole. If the dipole is placed in an external electric field of $4 \times 10^8 \ N/C$,calculate the maximum torque and the work done to rotate it by $180^{\circ}$.

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